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Did I solve this equation right? (Bernoulli)

  1. Feb 6, 2013 #1
    1. The problem statement, all variables and given/known data

    t^2y' + 2ty - y^3 = 0

    2. Relevant equations



    3. The attempt at a solution

    y'+(2/t)y = (1/t^2)y^3

    Let v = y^-2; then dv/dy = -2y^-3(dy/dt) and (dy/dt)=(-1/2)y^3(dv/dy)

    Making that sub we have:

    (-1/2)y^3(dv/dy)+(2/t)y = (1/t^2)y^3

    (dv/dy)+(-4/t)y^-2 = (-2/t^2)
    (dv/dy)+(-4/t)v = (-2/t^2)


    My integrating factor is t^-4, giving me

    v = (2/5)t^-1 + Ct^4
    y^-2 = (2/5)t^-1 + Ct^4
    y^2 = ((2/5)t^-1 + Ct^4)^-1

    y= (+/-) (((2/5)t^-1 + Ct^4)^-1)^(1/2)


    Wolfram gives something pretty irreconcilable.
     
  2. jcsd
  3. Feb 7, 2013 #2
    Your solution is correct. Note that you can plug in both values for y back into the DE to see if the left hand side equals 0, which is does.
     
  4. Feb 7, 2013 #3
    Thanks for the response, much appreciated.

    I'm having more trouble checking the validity of my solutions than finding them in the first place, turns into an algebraic mess.

    For y' = ry-ky^2 (k r constant) I get:

    y=((k/r)+ce^(-rt))^-1

    And again wolfram gives something weird, and for

    y' = ky - ny^3 I get

    y= (+/-)sqrt((n/k)+ce^(-2kt)^-1)

    Sorry again for the lack of latex, still doesn't work on my browser ( though admittedly if I knew the markup by memory without clicking the buttons it would work :/)
     
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